Question: Simplify and expand the following expression: $ \dfrac{r}{r + 1}-\dfrac{4r}{2r - 7} $
Explanation: In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(r + 1)(2r - 7)$ Multiply the first term by $\dfrac{2r - 7}{2r - 7}$ $ \begin{align*} \dfrac{r}{r + 1} \times \dfrac{2r - 7}{2r - 7} & = \dfrac{(r)(2r - 7)}{(r + 1)(2r - 7)} \\ & = \dfrac{2r^2 - 7r}{(r + 1)(2r - 7)}\end{align*} $ Multiply the second term by $\dfrac{r + 1}{r + 1}$ $ \begin{align*} \dfrac{4r}{2r - 7} \times \dfrac{r + 1}{r + 1} & = \dfrac{(4r)(r + 1)}{(2r - 7)(r + 1)} \\ & = \dfrac{4r^2 + 4r}{(2r - 7)(r + 1)}\end{align*} $ Now we have: $ = \dfrac{2r^2 - 7r}{(r + 1)(2r - 7)} - \dfrac{4r^2 + 4r}{(2r - 7)(r + 1)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{2r^2 - 7r - (4r^2 + 4r)}{(r + 1)(2r - 7)} $ $ = \dfrac{2r^2 - 7r - 4r^2 - 4r}{(r + 1)(2r - 7)} $ $ = \dfrac{-2r^2 - 11r}{(r + 1)(2r - 7)}$ Expand the denominator: $ = \dfrac{-2r^2 - 11r}{2r^2 - 5r - 7}$